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Chapter 7: Problem 79

Consider the stable elements through lead \((Z=82) .\) In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?

Short Answer

Expert verified
By creating a list of stable elements through lead \((Z=82)\) with their atomic weights and atomic numbers, we can compare the atomic weights and atomic numbers for each element sequentially. We count the mismatches where either the atomic weight of the current element is greater than that of its neighbor and its atomic number is less, or the atomic weight of the current element is less than that of its neighbor and its atomic number is greater. By following this process and counting the mismatches, we can determine the number of instances where atomic weights and atomic numbers are out of order relative to each other for stable elements up to lead \((Z=82)\).

Step by step solution

01

Prepare the list of stable elements

First, we ne to create a list of stable elements through lead (Z=82) with their atomic weights and atomic numbers. We'll use the periodic table to get this information for each element.
02

Compare atomic weights with atomic numbers

In this step, we'll compare the atomic weight and atomic number for each element. To do this, we go through the list, comparing the atomic weight of the current element with the atomic weight of its neighbor: - If the atomic weight of the current element is greater than that of its neighbor, and its atomic number is less than its neighbor's atomic number, we count it as a mismatch. - If the atomic weight of the current element is less than that of its neighbor, and its atomic number is greater than its neighbor's atomic number, we count it as a mismatch.
03

Count the total mismatches

We'll keep track of the mismatches as we go through the list of stable elements. When we reach the end of the list, the total number of mismatches will be our answer. The detailed list of comparisons may be too lengthy for this format, but by following these steps, we can find the number of instances where atomic weights and atomic numbers are out of order relative to each other for stable elements up to lead \((Z=82)\) in an orderly and efficient manner.

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Most popular questions from this chapter

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group 11 and group 12 metals are as follows: (a) Why are the electron affinities of the group 12 elements greater than zero? (b) Why do the electron affinities of the group 11 elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

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